Question

A circle with area A = pi r^2 is expanding with time. If dA/dt = 5cm^2/s, what is dr/dt when r=6?

Answer 1

Have you considered the 1st derivative in its differential form?

Answer 2

well I am asking because I don't know what to do?

Answer 3

Non-responsive.

You have \(A(r) = \pi r^{2}\). Can you find, from this definition, \(\dfrac{dA}{dr}\), the 1st derivative of A with respect to r?

Answer 4

is the derivative 2r?

Answer 5

Excellent. Put the \(\pi\) in there and you'll have it.

Answer 6

2pi r

Answer 7

or is it 2r pi?

Answer 8

*** Side Bar ***

\(A(r) = \pi r^{2}\)

\(\dfrac{dA}{dr} = 2\pi r\)

Did we just say the derivative of Area is the Circumference? I think we did!

*** End of Side Bar ***

Just for reference, it will help you if you try to use full and correct notation.

You have the derivative: \(\dfrac{dA}{dt} = 2\pi r\) This is the 1st Derivative in its "Functional Notation". Now, we do a little magic trick. It's kind of like multiplying by "dr", but it really isn't. It's just a different way to write it. \(dA = 2\pi r\;dr\). This is still the first derivative. It's just the "Differential Form". In this form, we just have a little algebra problem. Can you identify the required pieces?

\(dA = \) ??
\(dr = \) ??
\(r = \) ??

Answer 9

Multiplication is commutative. \(2\pi r\;and\;2r\pi\) are the same.

Answer 10

the first thing I want to do is divide by dr

Answer 11

Why would you do that? The problem statement asks you to find dr.

dA/dt = 5cm^2/s
r=6
\(\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}\)

Thus,

\(5 = 2\pi(6) \dfrac{dr}{dt}\) -- Solve for dt (or dr/dt, however you write it.)